Answer
$\log_{e}7.3891=2$
Work Step by Step
Since $b^y=x$ is equivalent to $y=\log_bx,$ the given equation, $
e^2=7.3891
,$ is equivalent to
\begin{array}{l}\require{cancel}
\log_{e}7.3891=2
.\end{array}
Note that $\log_ex$ is often abreviated $ln(x)$.
